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How do I argue that the closed ball of radius 1 isn't compact in $\mathbb{R}^{\infty}=\{ (x_1,x_2,\cdots) : x_i \in \mathbb{R} \ not \ all\ x_i \neq 0 \}$?

This set isn't metric space, so I can't use the fact that a compact set in a metric space is closed and bounded. Perhaps it isn't closed in $\mathbb{R}^{\infty}$.

Any hints?

Thanks in advance.

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    $\begingroup$ If this isn't a metric space, whats "a closed ball of raduis 1" then? $\endgroup$ – martini Oct 16 '13 at 8:08
  • $\begingroup$ In addition to defining "closed unit ball" you also have to specify what topology do you put on $R^{\infty}$ (there are many!). $\endgroup$ – Moishe Kohan Oct 16 '13 at 8:11
  • $\begingroup$ The topology here is $U \subset \mathbb{R}^{\infty} \ is \ open \ \Leftrightarrow U \cap \mathbb{R}^n \ is \ open \ in \ \mathbb{R}^n \ \forall n \in \mathbb{N}$. Your help is appreciated. $\endgroup$ – MathematicalPhysicist Oct 16 '13 at 8:30
  • $\begingroup$ Ok, then what do you mean by closed unit ball? $\endgroup$ – Moishe Kohan Oct 16 '13 at 16:35
  • $\begingroup$ $D=\{ x \in \mathbb{R}^{\infty} : ||x||=\sum x_i ^2 \leq 1 \}$ $\endgroup$ – MathematicalPhysicist Oct 16 '13 at 19:08
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First I’m going to fix the definitions:

$$\Bbb R^\infty=\{\langle x_k:k\in\Bbb Z^+\rangle:\exists m\in\Bbb Z^+\forall k\ge m(x_k=0)\}\;,$$

the set of all real sequences that are eventually $0$. For each $n\in\Bbb Z^+$ we define

$$R_n=\{\langle x_k:k\in\Bbb Z^+\rangle:x_k=0\text{ for all }k>n\}\;.$$

There is a natural bijection $\pi_n:R_n\to\Bbb R^n:\langle x_k:k\in\Bbb Z^+\rangle\mapsto\langle x_1,\ldots,x_n\rangle$. A set $U\subseteq\Bbb R^\infty$ is open iff $\pi_n[U\cap R_n]$ is open in the usual topology of $\Bbb R^n$. Finally, $$D=\left\{\langle x_k:k\in\Bbb Z^+\rangle\in\Bbb R^\infty:\sum_{k\ge 1}x_k^2\le 1\right\}\;,$$ and the problem is to prove that $D$ is not compact.


With that out of the way, here’s a large HINT:

For $n\in\Bbb Z^+$ let $x^{(n)}=\langle x_k^{(n)}:k\in\Bbb Z^+\rangle$, where

$$x_k^{(n)}=\begin{cases} 1,&\text{if }k=n\\ 0,&\text{otherwise}\;; \end{cases}$$

clearly each $x^{(n)}\in\Bbb R^\infty$. For each $n\in\Bbb Z^+$ let

$$U_n=\left\{\langle x_k:k\in\Bbb Z^+\rangle:\frac12<x_n<\frac32\text{ and }|x_k|<\frac12\text{ if }k\ne n\right\}\;.$$

  • Show that each $U_n$ is open in $\Bbb R^\infty$.

Now let $E=\{x^{(n)}:n\in\Bbb Z^+\}$.

  • Show that $E$ is closed in $\Bbb R^\infty$. Conclude that if $D$ were compact, $E$ would also be compact.
  • Show that $\{U_n:n\in\Bbb Z^+\}$ is an open cover of $E$ with no finite subcover, thereby showing that $E$ is not compact.
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  • $\begingroup$ I'll check this out tomorrow and after I'll understand it I'll approve it. Thanks. $\endgroup$ – MathematicalPhysicist Oct 17 '13 at 11:37
  • $\begingroup$ @MathematicalPhysicist: You’re welcome. $\endgroup$ – Brian M. Scott Oct 17 '13 at 18:24
  • $\begingroup$ In the definition of $R_n$ you meant that $x_k = 0 $, right? $\endgroup$ – MathematicalPhysicist Oct 19 '13 at 9:46
  • $\begingroup$ @MathematicalPhysicist: Oops. Yes, I sure did. Fixed; thanks. $\endgroup$ – Brian M. Scott Oct 19 '13 at 17:36

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